PHYS 611, SPR 12: HOMEWORK NO. 4 Solution Guide

Transcription

1 PHYS 611 SPR 12: HOMEWORK NO. 4 Solution Guide 1. In φ 3 4 theory compute the renormalized mass m R as a function of the physical mass m P to order g 2 in MS scheme and for an off-shell momentum subtraction scheme where we choose the constant c m so that Γ 2 ( m 2 ) =M 2 = M 2 m 2 R that is we demand that Σ(M 2 m 2 R) = 0. To solve: We start with the renormalized one-loop self energy in four dimensions Σ( m 2 g 2 { ( ) [ ] } 4πµ 2 β 1 ) = c 2(4π) 2 m γ E ln β ln (1) β + 1 m 2 R where β 1 4m2 R (2) and Σ is related to the 1PI two-point function Γ 2 by Γ 2 ( m 2 R) = m 2 R Σ( m 2 R). (3) At the physical mass Γ 2 (m 2 p m 2 R) = 0 so what we want to do is to solve this equation for m R in terms of m p. To do this in a practical way we use that Σ is already order g 2 and that m 2 R = m 2 P + O(g 2 ). This means that to order g 2 we can solve (3) by setting m 2 R = m 2 p inside Σ so that Γ 2 = 0 implies m 2 R = m 2 p Σ(m 2 p m 2 p). Then all we really need to do is figure out Σ(m 2 p m 2 p) for the (a) MS scheme and (b) momentum subtraction (MOM) scheme at = M 2. (a) MS scheme: Here c m = γ E ln 4π which gives Σ MS (m 2 p m 2 p) = { ( ) [ ] } g2 µ 2 β ln β 2(4π) 2 m 2 0 ln p β 0 + 1

2 where β 0 = 3 and we can readily check that the answer is real (see a previous HW set). (b) MOM scheme: In this case we have to determine c m from the requirement that Σ(M 2 m 2 R) = 0. A pretty short calculation then gives Σ MOM (m 2 p m 2 p) = g2 2(4π) 2 { β 0 ln [ ] β0 1 β β 1 ln [ ]} β1 1 β where now β 1 = 1 + 4m 2 P /M 2 contains the dependence on where we ve decided to normalize Σ. 2. Analyze the superficial degree of divergence for arbitrary 1PI diagrams in φ 4 4 and determine which classes of diagrams have UV divergences. To solve: Here it s just a matter of looking at the formula for the SDD in four dimensions ω(γ E ) = 4 E. ω 0 requires E 4. One- and two-point functions are not possible in φ 4. This is simple but not so obvious. One way to see it is from the symmetry φ φ of the Lagrange density. Any Green function with an odd number of external lines would correspond to a path integral that is odd under this change of variables and would thus vanish. Another way is to order the vertices in any diagram from left to right assigning some set of external lines as coming in from the left and the rest going out to the right. At each vertex the total number of lines is four so if an even (odd) number of lines comes into the vertex an even (odd comes out). Then if the total number of lines coming in is even (odd) the number of lines going out is also even (odd) and the number of lines coming in plus the number of lines going out will be even in either case. As a result the only choices for divergent diagrams are the self energy function E = 2 and the four-point scattering diagrams E = Renormalize φ 4 4 to one loop order in MS scheme. For the potential in the classical Lagrangian use the notation λ Rµ 2ε φ 4 4! R. (a) Verify that [λ R ] = 0 in D = 4 dimensions. (b) Show that only coupling constant and mass renormalization are necessary to this order and compute the renormalization constants Z m and Z λ to order λ R in a minimal subtraction scheme. To solve: (a) is just noting that in four dimensions [φ] = 1 so that d 4 x φ 4 is already dimensionless. The corresponding coupling must then also be dimensionless.

3 (b) To do this part we have to actually compute the one-loop diagrams with UV poles. By the previous homework we need only the one-loop corrections to the two- (E = 2) and three- (E = 3) point functions shown in the figures and identify (i) the ultraviolet poles and (ii) the counterterms that cancel these poles. The integrals themselves are pretty simple and have been given in class and in the notes. p 1 p p 1 p 4 1 p 4 p p 4 (a) (b1) (b2) (b3) Figure (a) can contribute only to mass renormalization because the diagram is independent of external momentum p. The integral for the diagram including the vertex but not the external lines is f(m R ) = ( iλ R )µ 2ε 1 2 d D q (2π) D i q 2 m 2 R + iɛ. Notice that the mass is the renormalized mass and the coupling the renormalized coupling inherited from the classical Lagrange density written in terms of renormalized quantities. We ve done the integral before and the pole term is just iλ R 1 2(4π) 2 m2 R ε. This is the terms we will cancel by a counterterm. Let s construct the counterterms after identifying the poles for the one-loop corrections to E = 4 Figs. (b-d). There are three integrals (p i +p j ) with (i j) = (1 2) (1 3) (1 4) which we can write as (p i + p j ) = ( iλ R ) 2 µ 4ε 1 2 = ( iλ R µ 2ε ) d D q (2π) D iλ R 1 2(4π) 2 ε + O(ε0 ) i (q 2 m 2 R + iɛ)((p i + p j ) 2 m 2 R + iɛ) and we note that the pole term is independent of the external momentum and is thus the same for all three diagrams. Notice also that we have factored out the lowest-order vertex iλ R µ 2ε because this pole should be thought of as a

4 modification to the vertex and this is the form of the counterterm. The other factors of µ ε contribute to finite terms. Now we can compute the counterterms which take the form (we suppress the subscripts R ) L ct = 1 2 ( µφ) 2 (Z φ 1) m2 2 φ2 (Z m Z φ 1) λ 4! φ4 (Z λ Z 2 φ 1). Because there are no UV poles proportional to in f( above there is no need for wave function renormalization at one loop and Z φ = 1 + O(λ 2 ). The counterterms we need are then found by direct comparison with the UV poles found above. To get the overall factors right we recall that a vertex of i(1/n!)φ n will give an overall factor of unity when the fields are replaced by variations with respect to sources in the path integral. The i s in f and above are the directly from the action extra factors of i associated with the variations are absorbed into the factors of i in the lines that connect to the vertex. In this way we conclude that the one-loop φ 4 counterterms are L (1) ct = 1 2 m2 φ 2 λ (4π) 2 ( 1 2ε + c λµ2ε 4! φ 4 λ (4π) 2 ( 3 2ε + c where the factor of 3 reflects the three 4-point diagrams with different flow of external momenta but the same UV pole. Since Z φ = 1 at one loop we can read off Z m = Z λ = ( λ 1 (4π) 2 2ε + c ( λ 3 (4π) 2 2ε + c. (4) 4. Compute the one-loop beta function for φ 4 4 solve for the running coupling and verify that the theory is not asymptotically free. To solve: Here we only need the relation between the beta function β(λ) = µdλ(µ)/dµ and the renormalization constant for the coupling Z λ given for D = 4 by β(λ) = 2λ 2 z 1(λ) λ.

5 where z 1 is the residue of the 1/ε pole in Z λ. To one loop this is found from the result of the previous problem as β(λ) = 3λ2 (4π) 2. Because the right-hand side is positive the coupling increases as µ increases and the theory is not asymptotically free.